# Superheated wing blast!

So, we know that big creatures will generate lots of heat if they’re active. We also know that wings are very good at venting excess heat. Now for most animals, this heat isn’t exactly spectacular, a few degrees here or there. Of course, if an animal generated more heat because it was really, really big, the effect might be a little more exciting. So let’s take a big creature that generates lots of heat. Let’s put big wings on this creature that vent the excess heat, and let’s say that this creature can control how the heat is vented from its wings. The creature vents the heat to the underside of its massive wings, creating large pockets of very hot air. Then, upon encountering something that the creature doesn’t like, it flaps its gargantuan wings and generates a hurricane force blast of scalding hot air.

So the question here is how much heat, and how much wind? For the purposes of this question, we’ll go big. Let’s say a 2-mile wingspan, each wing from 1/4 to 1 mile in breadth (base to tip), on a creature that is 2 miles long, with the heat generation you would expect from an active animal that size. If you can answer for both raptor-like wings and bat-like wings that would be great. For the purposes of this question ignore things like “how could this creature move” or “how could the wings withstand the force of being flapped” as those are different questions entirely.

Here’s a link to my other related questions 1: http://worldbuilding.stackexchange.com/questions/137872/the-nuclear-dragon

• We need to know the breadth of the wings, too (each wing is 1 mile wide, how high?) so we can calculate surface area. We also might need you to give us relative structure (the wings are built similar to what Terrestrial creature?) as the curve of the wings as they are brought to bear will affect the blast. Finally, we need to know exactly how much heat is being dissipated by the wings so we can calculate how quickly the air will heat. – JBH Jan 28 at 17:09
• I think your numbers are too big. The answers will probably come with a lot of buts and howevers. Maybe if you reduce your numbers to a kaiju-like scale you will get more feasible answers. – Magus Jan 28 at 19:45
• Nick, could you edit your posts in this series (preferably all of them) and cross-link them? Not only is that a benefit to other users, but SE tracks linked posts. (Maybe just link back to post #1, and keep a running list of all the questions in the series in that post). – JBH Jan 28 at 21:24
• Oh, and for fun and giggles I started to post a question on Physics about calculating the actual amount of air someone below a Robin would feel per-downstroke when I realized that almost nothing would be harder to integrate than the surface area of a bird's wing in flight, moment-by-moment, so that you can calculate actual air pressure and density under the wing. Mother Nature is very good at what she does. Cheers for a remarkably challenging question. – JBH Jan 28 at 21:26
• how strong is gravity? I doubt there's a substance known to man that can support a thin one mile cantilever in earth-like cravity. – Jasen Jan 29 at 3:38

## 3 Answers

A one mile long wing is big, I mean, REALLY big.

Let's say the "offensive wing beat" starts with the wing pointing backwards and completes 180 degress later pointing forwards.

The wing tip completes a full half circle with a one mile radius, so basically 3.14 miles.

How long does a beat take? 1 minute? 1 second? 10 seconds? Lets go with 10 seconds.

The wing tip will travel 3.14 miles in 10 seconds, or 1,884 miles per hour. The speed of sound - at sea level - is approx 770 miles per hour, so the wing tip is traveling at something like mach 2.5 during its rotation.

So for each beat the wings are displacing a couple of cubic miles of air in those 10 seconds, I'm guessing there'll be a massive pressure wave similar to a nuclear blast traveling forwards from your mega bird. The secondary blast is from the air rushing in to replace that moved by the wing beat, dragging debris with it.

I don't have enough physics to workout the actual forces involved, but the stored heat under the wings is the least of your worries.

• Sounds a bit like Randall Monroe :) – Cerberus Jan 29 at 4:26
• @Cerberus nah, if it was Randall, it'd be in comic form, with all the forces calculated ;) – Baldrickk Jan 29 at 9:18
• @Baldrickk: Absoutely! My high-school science was 30 years ago, Randal Monroe I am not :) – Binary Worrier Jan 29 at 9:27
• For some reason I can't stop laughing: Death by being kicked out from your own planet by a huge chicken gosh :) – Alma Do Jan 29 at 12:48
• Air works very differently at supersonic speeds from its behavior at subsonic speeds. I wonder if a supersonic wing flap even works... – KRyan Jan 29 at 15:15

## Edit

jdunlop has repeatedly pointed out the flaws of my answer so even if you have read the answer please reread it because there have been a few significant changes made. (in the Energy real world comparison part and following, it's a more accurate formula)

This simply is an expansion to Binary Worriers answer. (Math to display the true scale)

This is not that accurate because there are a lot of assumptions but it is good to display the ballpark in which the forces are working.

## Values:

• 1 mile (1,609.34 m) long wings
• 10 seconds to move 3.14 miles (ca. 5,053 m)
• we assume a 2:1 ratio in length to width so 0.5 miles (804.67 m) for our wing width
• this gets us a surface area of 1294987.62 $$m^2$$ or 0.5 $$miles^2$$
• A sphere of 1 mile (1,609.34 m) has a volume of 1.746 $$\cdot 10^{10}m^3$$ or 4.189 $$miles^3$$
• But our wing only has $$\frac {1}{4}$$ of the diameter the volume drops accordingly to 4 364 877 255 $$m^3$$ or 1.05 $$miles^3$$
• Lets reduce this by another 20% to adjust for imperfections of the wings and no full 180° flap of the wings (also wings don't tend to be perfectly shaped like the upper part of a circle). Which gives us 3 291 901 804 $$m^3$$ or 0.84 $$miles^3$$
• Finally we need the weight of air which is 1.275 kg per $$m^3$$

## Math

As Binary Worrier already explained the speed of each wing must be 1884 mph (3,032 km/h) to keep the numbers from exploding to astronomical scale we will assume that the wings already have this speed when starting to flap.

Weight

First we need the weight of the air which is our volume times the weight we already established. $$1.275 kg \cdot 3291901804 m^3 = 4,452,174,800 kg$$ This is 8.90435 TIMES THE WEIGHT OF THE BURJI KHALIFA. (The tallest building in the world currently)

Force

To get the force we use this trusty formula: $$F = m \cdot a;$$ F = force, m = mass, a = acceleration. Our acceleration is the speed of the wings divided by 10 (it takes 10 seconds for the wings to hit every air molecule in our designated volume) so 303.2 $$km/h^2$$ or 84.22222222 $$\frac {m}{s^2}$$.

So when plugged into our formula it looks like this: $$F = 4452174800 kg \cdot 84.22222 \frac {m}{s^2} = 3.75 \cdot 10^{11} N$$ (N = Newton)

Energy/Real world comparison (redone)

Energy is $$E = 0.5 \cdot m \cdot v^2$$ E = Energy, m = Mass, v = speed. Plugging our values in: $$E = 0.5 \cdot 4452174800 kg \cdot 842.22222222^2 \frac{m}{s} = 1.579 \cdot 10^{15}j$$ Which is about $$3 \cdot 10^{14}j$$ less but still enough to supply the world in 2015 for 6 days. Or 6 times less than a standard hurricane releases per second.

Edit V.2

For the creature unleashing the attack with the rough approximation for the weight by jdunlop we can estimate the creatures speed by the same formula we used above: $$E = 0.5 \cdot m \cdot v^2$$ this time we have to rearange it to get the speed: $$v^2 = \frac {E}{0.5 \cdot m}$$ again with our numbers: $$v^2 = \frac {1.579 \cdot 10^{15}j}{0.5 \cdot 20 \cdot 10^9 kg} = 157904.9$$ because v is squared we have to take the root and get: 397.37 m/s or 1430.54 km/h or 894.01 mph.

Conclusion (corrected weight comparison)

Anything that faces this creature has far greater problems than the heat released by this creature. The much bigger problem for the attacked creature is if it doesn't weigh more than 25 311 cars (each 1000 kg or 1 metric ton) it will leave an earth like planet for ever because it will reach escape velocity. And even when both (the attacked and the attacker) weigh more these creatures have to somehow not hit the ground, mountain, or hill with supersonic speed.

Additional Info

For the weight in cars needed not to leave the planet I used the formula from above and re arranged it this way: $$m = \frac {E}{0.5 \cdot v^2}$$ for the escape velocity is 11170 m/s. Couldn't use my physics book to site it but used its value instead of Wikipedia's the values are close enough though. So with the values: $$m = \frac {1.579 \cdot 10^{15}j}{0.5 \cdot 11170^2m/s} = 25310794 kg$$ or 25 310.794 metric tons.

• That final number looks suspiciously on the order of a nuclear blast.... But the analysis is missing the force of air resistance (the standing air that the cupped air is pushing against) forcing cupped air to spill over the wings. I suspect only a fraction of the air we're talking about would actually make it to the intended target. Nevertheless, +1 for a wonderful effort that does show the scale of the problem. – JBH Jan 28 at 20:36
• @JBH For simplicity I simply assumed that the lost air and the air resistance cancel each other out. But yeah for a higher accuracy this would be needed. – Soan Jan 28 at 20:42
• @Soan - I can see why you might doubt that, but that's not the case! Buildings are actually not very dense at all, whereas organic beings are. Take, for example, an unladen european swallow. Weight: 20g. Wingspan: 21cm. So a 10 cm (approximately) wing. To reach a mile long, we have to multiply that by a factor of 10 666 (or approximately 10000, in fermi numbers). Its volume (and therefore mass) would be multiplied by that factor cubed, and therefore would be twenty trillion grams, twenty billion kg. Burji Khalifa? A paltry 453 million kg. – jdunlop Jan 28 at 21:44
• (For a reference for which I have the data, the Willis tower masses about 202 million kg, and has a volume of approximately 1.4 million cubic metres, giving it a density of ~0.14 relative to water. Your average critter, even birds, has a density much closer to that of water.) – jdunlop Jan 28 at 21:51
• So, @Soan, my original comment was incorrect; a creature that big wouldn't weigh in the range of the BK - it would weigh much, much more. – jdunlop Jan 28 at 21:53

I agree with Binary Worrier's analysis, however: assuming this critter is in fact your 2-mile long nuclear dragon from an earlier question, the generation of massive amounts of heat under the wing surface helps answer the "how the hell does it FLY??" question very nicely - lift generation in a typical wing is driven by pressure differential between the lower straighter surface and the upper more curved surface of a wing (the air moves faster over the curved section and as a result, the 90 pressure against the wing itself is lessened, this more pressure beneath the wing) and secondarily through angle of attack dynamics, and in highly mobile wings on certain birds, root twist and other considerations. Your Nuke-dragon's excess heat radiators, in heating up the atmosphere beneath them, generate lift automagically, as the hotter gas expands it exerts more pressure on nearby surfaces.

Of course, the critter would need to cup the wings deeply to prevent escape of the heated gas... so the "flight" would be more like free ballooning - perhaps it's but one trick in the critter's flight methods - a way to do an equivalent to raising on thermals as hawks do absent an external thermal gradient - then your beastie closes wings behind and dives...